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Given vector M = (2cos & # 178; X, √ 3) n = (1, sin2x) function f (x) = m * n (2) In the triangle ABC, ABC is the opposite side of the angle ABC and f (c) = 3, C = 1 and a > b > C respectively
f(x)=m*n=2cos²x+√3sin2x
=1+cos2x+√3sin2x
=2(√3/2sin2x+1/2cos2x)+1
=2sin(2x+30)+1
f（C）=2sin(2C+30)+1=3
sin(2C+30)=1
2C+30=90
C=30,A+B=150
2R=c/sinC=1/1/2=2
√3a-b=2R(√3sinA-sinB)=2[3sinA-sin(150-A)]
=2[√3sinA-1/2cosA-√3/2sinA]
=2(√3/2sinA-1/2cosA)
=2sin(A-30)
thirty
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